1. ## Vector field

$F(x,y) = - \frac{y}{x^2+y^2}i + \frac{x}{x^2+y^2}j$

1) What domain of the field?

2) Prove that:
$\oint_{\alpha} - \frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = 2 \pi$

$\alpha$ is a circumference of center in origin and raio $r$

3) Check the green's theorem that:
$\oint_{\alpha} - \frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = 0$

My solution:

1) $IR- \{0,0\}$

2) $\int_0^{2 \pi} (sen^2t + cos^2t)dt = 2 \pi$

3) $\frac{ \partial (\frac{x}{x^2+y^2})}{ \partial x} - \frac{ \partial (\frac{y}{x^2+y^2})}{ \partial y} = 0$

$\int \int_S 0 dxdy = 0$

4) As the domain is $IR-\{0,0\}$ not is possible use the theorem of green

All this correct ?

2. Please it is correct ?

3. Originally Posted by Apprentice123
$F(x,y) = - \frac{y}{x^2+y^2}i + \frac{x}{x^2+y^2}j$

1) What domain of the field?

2) Prove that:
$\oint_{\alpha} - \frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = 2 \pi$

$\alpha$ is a circumference of center in origin and raio $r$

3) Check the green's theorem that:
$\oint_{\alpha} - \frac{y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy = 0$

My solution:

1) $IR- \{0,0\}$

2) $\int_0^{2 \pi} (sen^2t + cos^2t)dt = 2 \pi$

3) $\frac{ \partial (\frac{x}{x^2+y^2})}{ \partial x} - \frac{ \partial (\frac{y}{x^2+y^2})}{ \partial y} = 0$

$\int \int_S 0 dxdy = 0$

4) As the domain is $IR-\{0,0\}$ not is possible use the theorem of green

All this correct ?
Very good! If $\int_c P\,dx + Q\, dy = \iint_R Qx-Py \,dA$ then to use Green's theorem requires that $P$ and $Q$are continuous and have continuous partial derivatives in $R$ which as you stated is not true. As classic problem is to show that for your vector field that

$
\int_c P\,dx + Q\, dy = 2 \pi
$

for any simple closed curve C enclosing the origin.

4. thank you